poj 1080 Human Gene Functions 解题报告（附详细分析）

动态规划题，是最长子序列的变形。

最优解是字符串对应值的最大值，只需要改一下最长子序列的动态转移方程即可。

解题思路

1.取字符i-1和j-1的时候dp[i][j]=dp[i-1][j-1]+value[s1[i-1]][s2[j-1]];

2.取字符i-1不取j-1的时候dp[i][j]=dp[i-1][j]+value[s1[i-1]]['-'];

3.取字符j-1不取i-1的时候dp[i][j]=dp[i][j-1]+value['-'][s2[j-1]];

#include <iostream>
using namespace std;
const int MAX = 200;

int max(int, int, int);
int i, j;
int dp[MAX][MAX];
char s1[MAX], s2[MAX];
int value[MAX][MAX];

int main()
{
value['A']['A']=value['C']['C']=value['G']['G']=value['T']['T']=5;
value['A']['C']=value['C']['A']=value['A']['T']=value['T']['A']=-1;
value['-']['T']=value['T']['-']=-1;
value['A']['G']=value['G']['A']=value['C']['T']=value['T']['C']=-2;
value['G']['T']=value['T']['G']=value['G']['-']=value['-']['G']=-2;
value['A']['-']=value['-']['A']=value['C']['G']=value['G']['C']=-3;
value['C']['-']=value['-']['C']=-4;				//录入题目的表格
int t, m, n;
scanf("%d", &t);
while (t--)
{
scanf("%d%s", &n, s1+1);
scanf("%d%s", &m, s2+1);
memset(dp, 0, sizeof(dp));
dp[0][0] = 0;
for(i = 1; i <= n; i++)
dp[i][0] = dp[i-1][0] + value[s1[i]]['-'];	//数组边界的初始化
for(j = 1; j <= m; j++)
dp[0][j] = dp[0][j-1] + value['-'][s2[j]];	//当i=0时，表示第j个字母和‘-’匹配

for(i = 1; i <= n; i++)
for(j = 1; j <= m; j++)
{
dp[i][j] = max( dp[i-1][j-1] + value[s1[i]][s2[j]],
dp[i][j-1] + value['-'][s2[j]],		//表示s1[i]与s2[j-1]匹配，于是s2[j]就只能与‘-’匹配
dp[i-1][j] + value[s1[i]]['-'] );	//表示 s1[i-1] 与 s2[j] 匹配 于是 s1[i] 就只能和 ‘-’匹配
}
printf("%d\n", dp
[m]);
}
return 0;
}

int max(int a, int b, int c)
{
int m;
if(a > b)
m = a;
else
m = b;
if(m < c)
m = c;
return m;
}